Question

There were 49.7 million people with some type of long-lasting condition or disability living in the United States in 2000. This represented 19.3 percent of the majority of civilians aged five and over (http://factfinder.census.gov). A sample of 1000 persons is selected at random.

a. Approximate the probability that more than 200 persons in the sample have a disability.
b. Approximate the probability that between 180 and 300 people in the sample have a disability.

Answer 1

To approximate the probability that more than 200 persons in the sample have a disability, we calculate the mean and standard deviation of the binomial distribution and use the Z-table to find the probability. The approximate probability is 0.693. To approximate the probability that between 180 and 300 people in the sample have a disability, we calculate the z-scores for each value and find the probabilities using the Z-table. The approximate probability between 180 and 300 is 0.828.

Step-by-step explanation:

To approximate the probability that more than 200 persons in the sample have a disability, we need to use the normal approximation to the binomial distribution. First, we calculate the mean and standard deviation of the binomial distribution. The mean is given by np = 1000 * 0.193 = 193, and the standard deviation is given by sqrt(np(1-p)) = sqrt(1000 * 0.193 * 0.807) ≈ 13.84. To approximate the probability, we use the standard normal table (Z-table). We calculate the z-score for 200, which is (200 - 193) / 13.84 ≈ 0.505. Looking up the z-score in the Z-table gives a probability of approximately 0.693. Therefore, the approximate probability that more than 200 persons in the sample have a disability is 0.693.

To approximate the probability that between 180 and 300 people in the sample have a disability, we need to calculate the z-scores for each value. For 180, the z-score is (180 - 193) / 13.84 ≈ -0.940, and for 300, the z-score is (300 - 193) / 13.84 ≈ 7.68. Looking up these z-scores in the Z-table gives the probabilities of approximately 0.172 and essentially 1. The probability between 180 and 300 is the difference between these probabilities, which is approximately 1 - 0.172 ≈ 0.828. Therefore, the approximate probability that between 180 and 300 people in the sample have a disability is 0.828.

Answer 2

The answer is "0.274 and 0.841"

Step-by-step explanation:

Given:

In this question, let X the binomial random variable with the parameters:

`n = 1000\\\\P = 0.193`

Calculating the random variable Z:

`Z= (X - np)/(√(np(l - p)))=(x -193)/(12.48)`

is an approximately regular, cumulative random variable with
`\Phi` specified in the back of the book in the tables.

Using the continuity and tables to calculate:

`\to \mathbb{P} (X > 200) = \mathbb{P} (X\geq 200.5) \approx \mathbb{P}(Z\geq 0.6)=1- \Phi(0.6) \approx 0.274 \\\\\to \mathbb{P} (180 < X < 300) = \mathbb{P} (180.5 \leq X \leq 299.5) = \approx \mathbb{P} (-1 \leq Z\leq 8.53) = \approx 1- \Phi(-1) \approx 0.841`